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About LinkBackshmm are you sure?
when i did (X+Y)^2, i got
l4 0l
l0 4l
and i thought maybe A^n would be something like
l2^n-1 0l
l0 2^n-1l
that would work for all constants right?
No.You have computed it wrongly thenOriginally Posted by daRitz
First of all, X+Y = I
And $\displaystyle I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $
Try computing $\displaystyle I^n$ and tell me what you get...
$\displaystyle A = aX = \begin{pmatrix} a & a \\ a & a \end{pmatrix}$
Try computing $\displaystyle A^2, A^3$ and see....
Ok thanks i understand that one now!
Back to that previous problem,
A=aX, B=bX
you're formula was really good but i dont think that's what is supposed to be done because i didn't type the whole instructions:
Let A=aX and B=bX, where a and b are constants. use different velues of a and b to calculate A^2, A^3, A^4....(for references)
By considering integer powers of A and B, find expressions for A^n and B^n.
i honestly have no idea what they're asking in this problem, any idea?
"use different velues of a and b to calculate A^2, A^3, A^4....(for references)"
I am not sure what this means...
Probably your teacher wants you to guess the general formula for $\displaystyle A^n = (aX)^n$, by first trying some values.
Like you can try a=1, a=0 and a=2 and so on....
This will let you see a general pattern.... Then you guess that for a particular "a",
$\displaystyle A^n = \begin{pmatrix} a^n2^{n-1} & a^n2^{n-1} \\ a^n2^{n-1} & a^n2^{n-1} \end{pmatrix}$
Finally you should prove this guess using induction.......
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